National Math Olympiad

First, rewrite the equation as $3^3\cdot 3^{3\sin x}=(3^2{)}^{{\cos }^{2}x}$ and then $$3^{3+3\sin x}=3^{2{\cos }^{2}x}.$$ This implies ${\log }_3(3^{3+3\sin x})={\log }_3(3^{2{\cos }^{2}x})$ or $3+3\sin x=2{\cos }^{2}x$. Since ${\cos }^{2}x=1-{\sin }^{2}x$ we have $$1+3\sin x+2{\sin }^{2}x=0$$ and $$(1+\sin x)(1+2\sin x)=0.$$ This implies that $\sin x=-1$ or $\sin x=-1/2$. The only real number $x$ from the interval $[0,2\pi )$ satisfying the first condition is $x=\frac{3\pi }{2}$. There are two real numbers $x\in [0,2\pi )$, such that $\sin x=-1/2$, namely $x=\frac{7\pi }{6}$ and $x=\frac{11\pi }{6}$.

The solutions are $x=\frac{3\pi }{2}$, $x=\frac{7\pi }{6}$ and $x=\frac{11\pi }{6}$.